Answer

Verified

384.3k+ views

**Hint:**According to the problem there are $2n$ cards. The probability of drawing a number is proportional to $r$. The probability P(r) that the number $r$ is drawn in one draw is given by $P(r) = kr$.

Now, we find the probability P(r) for each number $1,2,3, \ldots ,2n(n \in N)$ is drawn in one draw by substituting $r = 1,2,3, \ldots ,2n(n \in N)$ into $P(r) = kr$.

From the postulate of probability that is the sum of the probabilities of all outcomes must equal 1, we find the value of k.

$ \Rightarrow P(1) + P(2) + P(3) + \ldots + P(2n) = 1$

Use the formula to find the sum of n natural number that is,

$1 + 2 + 3 + \ldots + n = \dfrac{{n(n + 1)}}{2}$

The sum of 2n natural number is obtained by substituting 2n into the formula of the sum of n natural number,

$1 + 2 + 3 + \ldots + 2n = n(2n + 1)$

In the given question we are asked to find the probability of drawing an even number in one draw.

we have to find the sum of probability P(r) for even numbers is drawn in one draw.

$ \Rightarrow P(2) + P(4) + P(6) + \ldots + P(2n)$

Substitute the value of k and sum of n natural numbers $1 + 2 + 3 + \ldots + n$ is $\dfrac{{n(n + 1)}}{2}$.

**Complete step-by-step solution:**

There are $2n$ cards, at each card the numbers printed as $1,2,3, \ldots ,2n(n \in N)$.

The probability of drawing a number is directly proportional to the $r$. That means is the probability P(r) that the number $r$ is drawn in one draw.

$ \Rightarrow P(r) = kr$

Where k is the proportionality constant.

Determine the probability for the values of $r$ as $1,2,3, \ldots ,2n(n \in N)$.

The probability of getting one is printed on the card is,

$P(1) = k(1)$

The probability of getting two is printed on the card is,

$P(2) = 2k$

The probability of getting three is printed on the card is,

$P(3) = 3k$

Similarly, the probability of getting $2n$ is printed on the card is,

$P(2n) = 2nk$

The sum of the probabilities of all outcomes must equal 1. $\therefore $The addition of the probability of getting $1,2,3, \ldots ,2n(n \in N)$ is equal to 1.

$ \Rightarrow P(1) + P(2) + P(3) + \ldots + P(2n) = 1$

$ \Rightarrow k + 2k + 3k + \ldots + 2nk = 1$

Take common term k outside from the left side of the equation,

$ \Rightarrow k(1 + 2 + 3 + \ldots + 2n) = 1 \ldots (1)$

Now, we can find the sum of 2n natural numbers.

Since the sum of n natural numbers $1 + 2 + 3 + \ldots + n$ is $\dfrac{{n(n + 1)}}{2}$.

To find sum of 2n natural numbers that is,$1 + 2 + 3 + \ldots + 2n$ is found by substituting n as 2n into the formula $\dfrac{{n(n + 1)}}{2}$.

$1 + 2 + 3 + \ldots + 2n = \dfrac{{2n(2n + 1)}}{2}$

$1 + 2 + 3 + \ldots + 2n = n(2n + 1) \ldots (2)$

Substitute the value of $n(2n + 1)$ into the equation (1),

$ \Rightarrow kn(2n + 1) = 1$

Solve the equation and find the value of k,

$ \Rightarrow k = \dfrac{1}{{n(2n + 1)}}$

According to the given question we are asked to find the probability of drawing an even number in one draw. So determine the addition of the probability for the even numbers,$ \Rightarrow P(2) + P(4) + P(6) + \ldots + P(2n)$

Since $P(r) = kr$ therefore,

$P(2) = 2k$

$P(4) = 4k$

$P(6) = 6k$

$ \Rightarrow P(2) + P(4) + P(6) + \ldots + P(2n)$

$ \Rightarrow 2k + 4k + 6k + \ldots + 2nk$

Take common term 2k outside from the left side of the equation,

$ \Rightarrow 2k(1 + 2 + 3 + \ldots + n)$

The sum of n natural numbers $1 + 2 + 3 + \ldots + n$ is $\dfrac{{n(n + 1)}}{2}$.

$ \Rightarrow 2k \times \dfrac{{n(n + 1)}}{2}$

Substitute $k = \dfrac{1}{{n(2n + 1)}}$ to get the probability of drawing an even number in one draw.

$ \Rightarrow 2 \times \dfrac{1}{{n(2n + 1)}} \times \dfrac{{n(n + 1)}}{2}$

Simplify the calculation,

$ \Rightarrow \dfrac{{n + 1}}{{2n + 1}}$

**Option D is the correct answer.**

**Note:**The most important step is to find the proportionality constant k. This can be found by the postulate, the sum of the probabilities of all outcomes must equal 1.

The formula that has to remember the sum of n natural numbers $1 + 2 + 3 + \ldots + n$ is $\dfrac{{n(n + 1)}}{2}$. By substituting 2n into the formula we get the sum of 2n natural numbers, $1 + 2 + 3 + \ldots + 2n$ is $\dfrac{{2n(2n + 1)}}{2}$ or $n(n + 1)$.

Recently Updated Pages

The base of a right prism is a pentagon whose sides class 10 maths CBSE

A die is thrown Find the probability that the number class 10 maths CBSE

A mans age is six times the age of his son In six years class 10 maths CBSE

A started a business with Rs 21000 and is joined afterwards class 10 maths CBSE

Aasifbhai bought a refrigerator at Rs 10000 After some class 10 maths CBSE

Give a brief history of the mathematician Pythagoras class 10 maths CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Name 10 Living and Non living things class 9 biology CBSE

Select the word that is correctly spelled a Twelveth class 10 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

Write the 6 fundamental rights of India and explain in detail